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If ln x + ln(x-7)= ln6x, then x= ?? show working out
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\[lnx+\ln(x-7)=lnx(x-7)\] so \[\ln(x^2-7x)=\ln(6x)\] If the logs are equal and the bases are the same, then the arguments must be equal.
\[x^2-7x=6x\] \[x^2-13x=0\] \[x(x-13)=0\] \[x=0, x=13\]
Discard x = 0
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